If sets $A, B$ are countable infinite then also $A \cup B$ This is my first „own“ proof I created. Is it ok? Are there things fishy? I‘d appreciate if someone could go thru my proof step by step and point out mistakes if there are any. $\Bbb N^{2n}$ means the even natural numbers and $\Bbb N^{2n+1}$ means the odd natural numbers.
Theorem: If sets $A, B$ are countable infinite then also $A \cup B$.
Proof: We assume that sets $A,B$ are countable infinite, i.e. $|A| = |\Bbb N| = |B|$, i.e. $f: A \to \Bbb N$ and $g: B \to \Bbb N$ are bijective.
We also know that $|\Bbb N^{2n}| = |\Bbb N| = |\Bbb N^{2n+1}|$.
It follows $|A| = |\Bbb N^{2n}|$ and $ |B| = |\Bbb N^{2n+1}|$, i.e. $f‘: A \to \Bbb N^{2n}$ and $g‘: B \to \Bbb N^{2n+1}$ are bijective.
Now we assume a function $h: A \cup B \to \Bbb N$ thru the rule: if $x \in A$ then $f‘(x)$, if $x \in B-A$ then $g‘(x)$.
The function $h$ is bijective. First it is injective because every preimage $x \in A \cup B$ has exactly one image due to the distribution of $f‘(x)$ and $g‘(x)$. It is also surjective because every image in $\Bbb N$ has exactly one preimage due to $f‘$, $g‘$.
But since $h$ is bijective it follows by definition $|A \cup B| = |\Bbb N|$ and that‘s by definition a case of countable infinity for $A \cup B$.
 A: Since you request in the comments for someone to specifically critique your proof, I'll do that here.

First, as mentioned, the notation $\Bbb N^{2n}$ for the even integer is highly unusual, and will certainly confuse your readers. As with numbers, one expects $\Bbb N^2$ to mean $\Bbb N\times \Bbb N$, and $\Bbb N^3$ to mean $\Bbb N\times \Bbb N\times\Bbb N$, and thus $\Bbb N^{2n}$ is usually seen as the cartesian product of $2n$ many copies of $\Bbb N$ (or alternatively as the set of functions from $2n=\{0,1,2,\dots,2n-1\}$ to $\Bbb N$).
To describe subsets unambiguously, one usually uses set builder notation, such as $\{2n\mid n\in\Bbb N\}$, which would be the set of even natural numbers. Some standard symbols for the even & odd numbers are $\Bbb E$ & $\Bbb O$, or $E$ & $O$ or $2\Bbb N$ & $2\Bbb N+1$.


Proof: We assume that sets $A,B$ are countable infinite, i.e. $|A| = |\Bbb N| = |B|$, i.e. $f: A \to \Bbb N$ and $g: B \to \Bbb N$ are bijective.

You mention $f$ and $g$, but they aren't used in the proof after this. I'd not state that $f$ and $g$ are bijective, since before this sentence these functions were still undefined. Instead, I'd say "hence, there exist bijective functions $f:A\to\Bbb N$ and $g:B\to \Bbb N$."

We also know that $|\Bbb N^{2n}| = |\Bbb N| = |\Bbb N^{2n+1}|$.
It follows $|A| = |\Bbb N^{2n}|$ and $ |B| = |\Bbb N^{2n+1}|$, i.e. $f‘: A \to \Bbb N^{2n}$ and $g‘: B \to \Bbb N^{2n+1}$ are bijective.

Same as before, you define $f`$ and $g`$. As far as symbols are concerned, you could consider using an apostrophe ' instead of a backtick `.

Now we assume a function $h: A \cup B \to \Bbb N$ thru the rule: if $x \in A$ then $f‘(x)$, if $x \in B-A$ then $g‘(x)$.

We don't assume $h$, we define it. Also, "then $f`(x)$" doesn't mean anything, you should state "then $h(x)=f`(x)$" instead.

The function $h$ is bijective. First it is injective because every preimage $x \in A \cup B$ has exactly one image due to the distribution of $f‘(x)$ and $g‘(x)$. It is also surjective because every image in $\Bbb N$
has exactly one preimage due to $f‘$, $g‘$.
But since $h$ is bijective it follows by definition $|A \cup B| = |\Bbb N|$ and that‘s by definition a case of countable infinity for $A \cup B$.

This is arguably the crux of the proof, but unfortunately also where the proof is most wrong. "Due to" is not exactly precise enough, and indeed it is even incorrect in the argument of surjectivity.
The reason that no two $x,x'\in A\cup B$ map to the same element $h(x)=h(x')$ is because of three facts:

*

*$h\restriction A=f`$ is a bijection, and thus injective,

*$h\restriction (B-A)=g`\restriction (B-A)\subseteq g`$, and $g`$ is injective since $g`$ is a bijection,

*the range of $f`$ (the even numbers) and of $g`$ (the odd numbers) are disjoint.

This final point is crucially why the composite function $h$ is injective.
As far as surjectivity of $h$ is concerned, consider the following $A$, $B$, $f`$ and $g`$:

*

*$A=\{n\in\Bbb N\mid n\geq 1\}=\{1,2,3,4,5,6,\dots\}$,

*$B=\Bbb N=\{0,1,2,3,4,\dots\}$,

*$f`:n\mapsto 2n-2$

*$g`:n\mapsto 2n+1$
Now the range of $h$ includes the range of $f`$ and thus includes all even numbers, but it does not include all odd numbers, since only $g`(0)=1$ is in the range of $h$. This is because $B-A=\{0\}$ contains only one element. So $h$ is not surjective. In fact, the way you defined $h$, we will only get surjectivity if $A$ and $B$ are disjoint sets.
However, this is actually not a problem: you don't have to give an explicit bijection between $A\cup B$ and $\Bbb N$. Instead, since $h$ is injective, you've already proved that $|A\cup B|\leq |\Bbb N|$, so it is enough to now find an injective function $\Bbb N\to A\cup B$. But this is easy, since $f:A\to \Bbb N$ is bijective, thus $f^{-1}:\Bbb N\to A$ fits our purpose.
Since $|A\cup B|\leq |\Bbb N|$ and $|\Bbb N|\leq |A\cup B|$, we have $|A\cup B|=|\Bbb N|$ by the Cantor-Schröder-Bernstein theorem (note that this theorem is not as trivial as it looks symbolically).

If you don't want to use the Cantor-Schröder-Bernstein theorem, I suggest to split your $A\cup B$ into three sets: $X_1=A-B$, $X_2=B-A$ and $X_3=A\cap B$. You can then make a case distinction for which of the $X_i$ is finite and which of the $X_i$ is infinite. They can't all be finite by the pigeonhole principle. This way you can write down an explicit bijection as well, but it's a bit more work.
